The radius of a right circular cylinder increases at the rate of 0.2 cm/sec and the height decreases at the rate of 0.1 cm/sec. The rate of change of the volume of the cylinder when the radius is 1 cm and the height is 2 cm is

Options

A. `(pi)/(10)cm^(3)//sec`
B. `(9pi)/(10)cm^(3)//sec`
C. `(7pi)/(10)cm^(3)//sec`
D. `(pi)/(5)cm^(3)//sec`

Answer

To find the rate of change of the volume of a right circular cylinder when the radius is 1 cm and the height is 2 cm, we can follow these steps:

### Step 1: Write down the formula for the volume of a cylinder.
The volume \( V \) of a right circular cylinder is given by the formula:
\[ V = \pi r^2 h \]

### Step 2: Differentiate the volume with respect to time \( t \).
Using the product rule and the chain rule, we differentiate \( V \):
\[ \frac{dV}{dt} = \pi \left( 2rh \frac{dr}{dt} + r^2 \frac{dh}{dt} \right) \]

### Step 3: Substitute the known values.
We know:
- \( r = 1 \) cm
- \( h = 2 \) cm
- \( \frac{dr}{dt} = 0.2 \) cm/sec (rate of increase of radius)
- \( \frac{dh}{dt} = -0.1 \) cm/sec (rate of decrease of height)

Substituting these values into the differentiated volume equation:
\[ \frac{dV}{dt} = \pi \left( 2 \cdot 1 \cdot 2 \cdot 0.2 + 1^2 \cdot (-0.1) \right) \]

### Step 4: Calculate the terms inside the parentheses.
Calculating each term:
1. \( 2 \cdot 1 \cdot 2 \cdot 0.2 = 0.8 \)
2. \( 1^2 \cdot (-0.1) = -0.1 \)

Now substitute these results back into the equation:
\[ \frac{dV}{dt} = \pi (0.8 - 0.1) = \pi (0.7) \]

### Step 5: Final calculation.
Thus, we have:
\[ \frac{dV}{dt} = 0.7\pi \text{ cm}^3/\text{sec} \]

### Step 6: Conclusion.
The rate of change of the volume of the cylinder when the radius is 1 cm and the height is 2 cm is:
\[ \frac{dV}{dt} = \frac{7\pi}{10} \text{ cm}^3/\text{sec} \]

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